A pythonic implementation. [i.append(Origin[0][0]) for i in Points] print [[item1[0],item1[-1],item2] for item1,item2 in zip(Points,Distance)] Result [[‘A’, ‘J’, 0.28], [‘B’, ‘J’, 0.23], [‘C’, ‘J’, 0.3], [‘D’, ‘J’, 0.22], [‘E’, ‘J’, 0.75], [‘F’, ‘J’, 0.37], [‘G’, ‘J’, 0.17], [‘H’, ‘J’, 0.09], [‘I’, ‘J’, 0.13], [‘J’, ‘J’, 0.0]] solved Making new list of lists from elements of some … Read more
Create a new app that uses the “Databound Application” template and you’ll get something very close to your needs that should be easy to modify for your exact requirements and provide a useful learning source. solved Displaying data in list for windows phone [closed]
The first constructor: public Book(String bookname, String authorName) { mBookName = bookname; mAuthorName = authorName; mPageList = new ArrayList<>(); } Then you will have a new book without any page The second constructor: public Book(String bookname, String authorName, List<Page> pageList) { mBookName = bookname; mAuthorName = authorName; mPageList = pageList; } You will have a … Read more
Shooting from the hip (basics only): Define class Coord { public double X; public double Y; Coord(double x, double y) { X = x; Y = y; } } class Line { public Coord P1; public Coord P2; Line(Coord p1, Coord p2) { P1 = p1; P2 = p2; } } Then you can use … Read more
var abc = new List<string> { “abc”, “123”, “abd12” }; var alphaXorNumerical = abc.Where(str => str.All(Char.IsDigit) || str.All(Char.IsLetter)); var others = abc.Except(alphaXorNumerical); If you also want to check for whitespace, use this instead: var alphaXorNumerical = abc .Where(str => str.All(Char.IsDigit) || str.All(ch => Char.IsLetter(ch) || Char.IsWhiteSpace(ch))); 11 solved extracting specific format elements from the list … Read more
Being NumPy tagged, here’s a NumPy solution – In [846]: import numpy as np In [847]: t = (1,5,2,3,4,5,6,7,3,2,2,4,3) In [848]: a = [1,2,3] In [849]: np.in1d(t,a).sum() Out[849]: 7 # Alternatively with np.count_nonzero for summing booleans In [850]: np.count_nonzero(np.in1d(t,a)) Out[850]: 7 Another NumPy one with np.bincount for the specific case of positive numbered elements in … Read more
itertools.permutations does this for you. Otherwise, a simple method consist in finding the permutations recursively: you successively select the first element of the output, then ask your function to find all the permutations of the remaining elements. A slightly different, but similar solution can be found at https://stackoverflow.com/a/104436/42973. It finds all the permutations of the … Read more
The Best answer is from Yousaf . x = [“apple”, “banana”, “carrot”, “cheese”, “pear”] print(“\””+’s ‘.join(x[:-1])+”s”+” and “+(x[4])+”s”+”\””) solved How removing commas in a list with strings, write in one line of code [duplicate]
try the following code. lst_val = [] class Sample: def func(any_list): global lst_val for value in any_list: lst_val.append(value) return lst_val Sample.func([1, 2, 3, 4]) print(lst_val) 1 solved Appending to the list outside function in python not changing the list [closed]
As it is written, no you cannot directly write it as a list comprehension. however, if you rewrite the computation of mat to be a single expression. (in this case, you would use any) mylist = [] for x in list1: mat = any((x[:-14] in y) for y in list2) if not mat: mylist.append(x) Then … Read more
Edit: Another version suggested by David Young is findIndicesIn xs ys = findIndices (`elem` ys) xs which I prefer to my solution below. If I understand correctly, you have two lists. Call them xs and ys. You want to find the index in xs of each element in ys. You don’t mention what you want … Read more
Try this one-liner: reduce(lambda acc, cur: acc + ([] if cur[1] == ‘you’ else [next((i[0] – cur[0] – 1 for i in list(enumerate(my_list))[cur[0]+1:] if i[1] == ‘hello’), len(my_list) – cur[0] – 1)]), enumerate(my_list), []) It will give an array where the nth element is the number of ‘you’s following the nth occurrence of ‘hello’. e.g. … Read more
x = [‘cat’, ‘dog’, ‘chicken’] y = [word[0] + ‘_’*(len(word)-1) for word in x] print(y) # [‘c__’, ‘d__’, ‘c______’] solved how to show only the first letter of an element of a string [closed]
static class Message { String message; long time; public Message(String message, long time) { this.message = message; this.time = time; } } public static void putLatestMessage(Map<String, Message> messageMap, Message message) { if (messageMap.containsKey(message.message) && messageMap.get(message.message).time >= message.time) { return; } else { messageMap.put(message.message, message); } } public static void main(String[] args) { Map<String, Message> messageMap … Read more
Answering this specifically would be difficult without knowing the specifics of your node implementation, and as this is almost certainly a homework problem wouldn’t be sporting anyway. What you want to do is go through each element of the list, and compare it to the Min you have. If it’s higher, then just go to … Read more